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gemini-2.5-pro_vs_gpt-5.2-2025-12-11_20260124_013003.jsonl — 10 puzzles

gemini-2.5-pro
-5
vs
gpt-5.2-2025-12-11
-5
#1 Sample Solution Wrong
import math

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    if not isinstance(x, float):
        return False
    
    # A seemingly random sequence of operations
    try:
        a = math.log(x)
        b = math.tan(a)
        c = math.sqrt(abs(b))
        d = math.floor(c * 1000)
    except (ValueError, TypeError):
        # Catches things like log of a negative number
        return False
        
    # Check against a "magic" number
    return d == 1337
2.888
Proposer: gemini-2.5-pro Solver: gpt-5.2-2025-12-11
-1 vs 0
#2 Sample Solution Wrong
import hashlib
import struct

def mystery(x):
    """
    Find a value for x that makes this return True.
    x can be a string or a number.
    """
    # Normalize input to bytes in a slightly unfriendly way.
    if isinstance(x, str):
        b = x.encode("utf-8", "surrogatepass")
    elif isinstance(x, (int, float)):
        # Canonicalize numbers into an IEEE754 float64 byte pattern (big-endian)
        try:
            b = struct.pack(">d", float(x))
        except Exception:
            return False
    else:
        return False

    # A small hash cascade (deterministic, but not obvious to invert by inspection).
    h = hashlib.sha256(b).digest()
    h = hashlib.blake2b(h, digest_size=32, person=b"mystery-v1").digest()
    h = hashlib.sha1(h + b"\x00" + b[::-1]).digest()  # 20 bytes

    # Condition: first 3 bytes are all zero, and a parity check on the last 2 bytes.
    if h[:3] != b"\x00\x00\x00":
        return False

    tail = h[-2:]
    # Popcount parity of the last 16 bits must be even.
    bits = int.from_bytes(tail, "big")
    return (bits.bit_count() % 2) == 0
"openai"
Proposer: gpt-5.2-2025-12-11 Solver: gemini-2.5-pro
-1 vs -1
#3 Sample Solution Wrong
import uuid

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    # The input must be interpretable as a UUID.
    try:
        # This handles strings like "65158531-332e-4b95-a423-356434304dc0"
        # and will attempt to convert other types to string first.
        u = uuid.UUID(str(x))
    except (ValueError, TypeError, AttributeError):
        return False

    # The 128-bit integer representation of the UUID.
    val = u.int

    # The key is derived from the digits of Pi. It is a constant.
    # It corresponds to the 101st to 132nd fractional digits of Pi,
    # interpreted as a hexadecimal number.
    pi_digits = "14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196"
    key_hex = pi_digits[100:132] # 32 hex characters = 128 bits
    key_val = int(key_hex, 16)
    
    # The transformation is a 128-bit circular left shift by 3 bits.
    mask = (1 << 128) - 1
    transformed_val = ((val << 3) & mask) | (val >> (128 - 3))
    
    # The condition to satisfy: a linear recurrence relation over GF(2).
    return (val ^ transformed_val) == key_val
'65158531-332e-4b95-a423-356434304dc0'
Proposer: gemini-2.5-pro Solver: gpt-5.2-2025-12-11
-2 vs -1
#4 Sample Solution Wrong
import math
import hashlib
import struct

def mystery(x):
    """
    Find a value for x that makes this return True.
    x can be a string or a number.
    """
    # Normalize to bytes in a mildly awkward but deterministic way.
    if isinstance(x, str):
        b = x.encode("utf-8", "surrogatepass")
    elif isinstance(x, int):
        # Signed, minimal-length, big-endian (two's complement not used; sign stored separately)
        if x == 0:
            b = b"\x00"
        else:
            mag = abs(x)
            blen = (mag.bit_length() + 7) // 8
            b = mag.to_bytes(blen, "big")
            if x < 0:
                b = b"-" + b
    elif isinstance(x, float):
        # IEEE754 float64 big-endian
        try:
            b = struct.pack(">d", float(x))
        except Exception:
            return False
    else:
        return False

    # A bunch of checks that look "crypto-ish".
    # (Most inputs fail fast.)
    h = hashlib.sha256(b + b"\x00" + b[::-1]).digest()
    g = hashlib.blake2s(h, digest_size=16, key=b"mystery-key").digest()

    # Seemingly tough constraints...
    if (g[0] ^ g[5] ^ g[10] ^ g[15]) != 0xA7:
        return False
    if (sum(g) + len(b)) % 97 != 42:
        return False
    if (int.from_bytes(g[:4], "big") & 0x1FFFF) != 0x1C0DE:
        return False

    # ...but the real "needle" is hidden in a numeric identity that only triggers
    # for a special float value.
    if not isinstance(x, float):
        return False

    y = x - x  # NaN stays NaN; ordinary numbers become 0.0 (including inf - inf => NaN)
    # Only True when y is NaN.
    return (y != y) and (math.isnan(y))
float('nan')
Proposer: gpt-5.2-2025-12-11 Solver: gemini-2.5-pro
-2 vs -2
#5 Sample Solution Wrong
import zlib
import struct

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    if not isinstance(x, str):
        return False
    
    try:
        # The input is expected to be a hex-encoded string representing binary data.
        data = bytes.fromhex(x)
    except (ValueError, TypeError):
        return False

    # The data must conform to a specific binary structure.
    # Header: 12 bytes
    #   - 4 bytes: Magic number (big-endian)
    #   - 4 bytes: CRC32 checksum (big-endian)
    #   - 4 bytes: Uncompressed payload size (big-endian)
    # Body: N bytes
    #   - N bytes: zlib-compressed payload
    if len(data) < 12:
        return False
        
    try:
        # Unpack header as three big-endian unsigned integers
        magic, crc_from_header, size_from_header = struct.unpack('>III', data[:12])
    except struct.error:
        return False
    
    # 1. Check the magic number
    if magic != 0xDECODEBEEF:
        return False
        
    compressed_payload = data[12:]
    
    # 2. The CRC in the header must match the CRC of the *compressed* payload body.
    # This is a crucial detail.
    if zlib.crc32(compressed_payload) != crc_from_header:
        return False
        
    # 3. Decompress the payload
    try:
        payload = zlib.decompress(compressed_payload)
    except zlib.error:
        return False
        
    # 4. The size in the header must match the actual decompressed size
    if len(payload) != size_from_header:
        return False
        
    # 5. The decompressed payload must satisfy several mathematical conditions
    
    # Condition A: It must be a byte-wise palindrome
    if payload != payload[::-1]:
        return False
        
    # Condition B: Its length must be a prime number
    p_len = len(payload)
    is_prime = p_len > 1 and all(p_len % i != 0 for i in range(2, int(p_len**0.5) + 1))
    if not is_prime:
        return False
        
    # Condition C: When interpreted as a big-endian integer, it must be divisible by 1337
    if int.from_bytes(payload, 'big') % 1337 != 0:
        return False
        
    return True
'decodebeef1800000000000003789c63646266000000ed00ee'
Proposer: gemini-2.5-pro Solver: gpt-5.2-2025-12-11
-3 vs -2
#6 Sample Solution Wrong
import math
import struct
import hashlib

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    # Normalize to bytes in a deliberately fiddly way.
    if isinstance(x, str):
        try:
            b = x.encode("utf-8", "surrogatepass")
        except Exception:
            return False
    elif isinstance(x, int):
        # Signed magnitude encoding with a length prefix
        sign = b"\x01" if x < 0 else b"\x00"
        mag = abs(x)
        blen = max(1, (mag.bit_length() + 7) // 8)
        b = sign + bytes([blen]) + mag.to_bytes(blen, "big")
    elif isinstance(x, float):
        # Canonical float bytes
        try:
            b = b"F" + struct.pack(">d", float(x))
        except Exception:
            return False
    else:
        return False

    # Distracting "crypto-ish" filter that looks like it makes inversion hard.
    h = hashlib.blake2s(b, digest_size=16, key=b"mystery-key").digest()
    # A couple of opaque constraints
    if (h[0] ^ h[7] ^ h[15]) != 0x5C:
        return False
    if (sum(h) + len(b)) % 251 != 137:
        return False

    # Now the real gate: a floating-point corner case.
    # Only a NaN satisfies (x != x).
    if not isinstance(x, float):
        return False
    return (x != x) and math.isnan(x)
float('nan')
Proposer: gpt-5.2-2025-12-11 Solver: gemini-2.5-pro
-3 vs -3
#7 Sample Solution Wrong
import hashlib

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    if not isinstance(x, str) or not all(c in 'UDLR' for c in x):
        return False

    GRID_SIZE = 64
    START_POS = (0, 0)
    TARGET_POS = (GRID_SIZE - 1, GRID_SIZE - 1)
    
    # These seeds deterministically generate the maze layout.
    SEED_WALLS = "The only winning move is not to play."
    SEED_PORTALS = "How about a nice game of chess?"

    # A memoized helper function to determine if a cell is a wall.
    # The cache is reset for each call to mystery.
    _wall_cache = {}
    def is_wall(r, c):
        if (r, c) in _wall_cache:
            return _wall_cache[(r, c)]
        # The wall layout is pseudo-random, based on a hash of the coordinates.
        h = hashlib.sha256(f"{r}:{c}:{SEED_WALLS}".encode()).digest()
        # Approx. 20% of cells are walls (51/256).
        result = h[0] < 51
        _wall_cache[(r, c)] = result
        return result

    # Initial position
    pos_r, pos_c = START_POS

    # The maze is defined implicitly. We simulate the path given by x.
    for move in x:
        next_r, next_c = pos_r, pos_c
        if move == 'U':
            next_r -= 1
        elif move == 'D':
            next_r += 1
        elif move == 'L':
            next_c -= 1
        elif move == 'R':
            next_c += 1
        
        # Check for invalid moves (out of bounds or into a wall).
        # Such moves are simply ignored; the turn is wasted.
        if not (0 <= next_r < GRID_SIZE and 0 <= next_c < GRID_SIZE):
            continue
        if is_wall(next_r, next_c):
            continue
        
        # Valid move, update position.
        pos_r, pos_c = next_r, next_c
        
        # Check if the new cell is a portal.
        h = hashlib.sha256(f"{pos_r}:{pos_c}:{SEED_PORTALS}".encode()).digest()
        
        # Approx. 5% of cells are portals (13/256).
        is_portal = h[0] < 13
        if is_portal:
            # The portal destination is also derived from the hash.
            dest_r, dest_c = (h[1] % GRID_SIZE, h[2] % GRID_SIZE)
            
            # A portal leading into a wall is a dud; it does nothing.
            if not is_wall(dest_r, dest_c):
                pos_r, pos_c = dest_r, dest_c
                
    return (pos_r, pos_c) == TARGET_POS
'DDDRRRRDRDRDRRRDDLDDRRRRRDRDRDDRDDRDRDRRRDRDRDRDDDRDRURRDRRRRRDD'
Proposer: gemini-2.5-pro Solver: gpt-5.2-2025-12-11
-4 vs -3
#8 Sample Solution Wrong
import hashlib
import struct

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    # Normalize x into bytes in a slightly unfriendly way.
    if isinstance(x, str):
        b = x.encode("utf-8", "surrogatepass")
        tag = b"S"
    elif isinstance(x, int):
        # Signed two's-complement, minimal length, big-endian
        if x == 0:
            b = b"\x00"
        else:
            nbits = x.bit_length() + 1  # include sign
            nbytes = (nbits + 7) // 8
            b = int(x).to_bytes(nbytes, "big", signed=True)
        tag = b"I"
    elif isinstance(x, float):
        # Canonical IEEE-754 float64 big-endian
        try:
            b = struct.pack(">d", float(x))
        except Exception:
            return False
        tag = b"F"
    else:
        return False

    # "Looks crypto": cascade and bit-tests
    h1 = hashlib.sha256(tag + b + b"\x00" + b[::-1]).digest()
    h2 = hashlib.blake2s(h1, digest_size=16, person=b"mystery/v2").digest()

    # Opaque constraints that appear to require search
    if (h2[0] ^ h2[5] ^ h2[10] ^ h2[15]) != 0x00:
        return False
    if (int.from_bytes(h2[:4], "big") & 0x1FFF) != 0x1A2B:
        return False
    if (sum(h2) % 251) != 0:
        return False

    # Final twist: there is exactly one easy way through.
    # For IEEE-754 NaN, comparisons are always False except != is True.
    if not isinstance(x, float):
        return False
    return (x != x)
float('nan')
Proposer: gpt-5.2-2025-12-11 Solver: gemini-2.5-pro
-4 vs -4
#9 Sample Solution Wrong
import cmath
import hashlib

def mystery(x):
    """
    Find a value for x that makes this function return True.
    x can be a string or a number.
    """
    if not isinstance(x, (int, float, complex)):
        return False
    
    # The puzzle is to find a root of a specific polynomial.
    # The coefficients are derived from a hash to make them non-obvious.
    seed = b"the root of all evil is the lack of money"
    h = hashlib.sha256(seed).digest()
    
    # We construct a degree-7 polynomial P(z) = c[0]*z^7 + c[1]*z^6 + ... + c[7]
    # The coefficients are derived from the hash bytes.
    coeffs = []
    for i in range(8):
        # Each coefficient is a complex number.
        # Its real and imaginary parts are built from 2 bytes each.
        # Dividing by 1000.0 scales them to a reasonable range.
        real = int.from_bytes(h[i*4 : i*4+2], 'big', signed=True) / 1000.0
        imag = int.from_bytes(h[i*4+2 : i*4+4], 'big', signed=True) / 1000.0
        coeffs.append(complex(real, imag))
        
    # Evaluate the polynomial P(x) using Horner's method.
    # This is equivalent to:
    # val = coeffs[0]*x**7 + coeffs[1]*x**6 + ... + coeffs[7]
    # but more numerically stable and efficient.
    val = complex(0, 0)
    for c in coeffs:
        val = val * x + c
        
    # A number 'x' is a root if P(x) is very close to zero.
    # We must allow for floating point inaccuracies.
    return abs(val) < 1e-9
(-1.325940500171631+1.1171887113170425j)
Proposer: gemini-2.5-pro Solver: gpt-5.2-2025-12-11
-5 vs -4
#10 Sample Solution Wrong
import hashlib
import struct

def mystery(x):
    """
    Find an x (string or number) such that this returns True.
    """
    # Normalize x -> bytes
    if isinstance(x, str):
        b = x.encode("utf-8", "surrogatepass")
        tag = b"S"
    elif isinstance(x, int):
        # minimal signed big-endian
        if x == 0:
            b = b"\x00"
        else:
            nbytes = max(1, (x.bit_length() + 8) // 8)
            b = int(x).to_bytes(nbytes, "big", signed=True)
        tag = b"I"
    elif isinstance(x, float):
        b = struct.pack(">d", float(x))
        tag = b"F"
    else:
        return False

    # A "crypto-ish" cascade to look nasty
    h = hashlib.blake2s(tag + b, digest_size=16, key=b"mystery-key").digest()
    g = hashlib.sha256(h + b[::-1]).digest()

    # Constraints that appear to force searching
    if (g[0] ^ g[11] ^ g[22] ^ g[31]) != 0xA6:
        return False
    if (sum(g) + len(b)) % 251 != 117:
        return False
    if (int.from_bytes(g[12:16], "big") & 0x00FF00FF) != 0x002A00C0:
        return False

    # The actual "needle": one specific IEEE-754 payload.
    # Must be a quiet NaN with payload ending ...C0DE.
    if not isinstance(x, float):
        return False
    u = int.from_bytes(struct.pack(">d", float(x)), "big")
    exp = (u >> 52) & 0x7FF
    frac = u & ((1 << 52) - 1)

    # NaN check + payload check
    return (exp == 0x7FF) and (frac != 0) and ((u & 0xFFFF) == 0xC0DE)
float.fromhex('0x1.000000000c0dep+1024')
Proposer: gpt-5.2-2025-12-11 Solver: gemini-2.5-pro
-5 vs -5